Question

A father says to son, "Six years before your birth, I w as twice as old as you mother. Ten years ago your mother was twice as old you then were. Three year hence I shall be twice as old as you will be. Find the present ages of the three.

Answer 1

Let the age of father, mother, and son be $x,y$ and $z$ respectively.

Now according to the question-

Condition I:-

$x-(z+6)=2(y-(z+6)]$

$x-z-6=2(y-z-6)$

$x-z-6=2y-2z-12$

$x-2y+z+6=0.....\left(1\right)$

Condition II:-

$y-10=2(z-10)$

$y-10=2z-20$

$\Rightarrow y=2z-10.....\left(2\right)$

Condition III:-

$x+3=2(z+3)$

$x+3=2z+6$

$x=2z+3.....\left(3\right)$

Now, fro ${eq}^n\left(1\right),\left(2\right)\&\left(3\right)$, we have

$(2z+3)-2(2z-10)+z+6=0$

$2z+3-4z+20+z+6=0$

$29-z=0$

$\Rightarrow z=29$

Substituting the value of $z$ in ${eq}^n\left(2\right)\&\left(3\right)$, we get

$y=2\times 29-10=48$

$x=2\times 29+3=61$

Hence the age of father, mother and son is $61$years, $48$ years and $29$ years, respectively.